6. Length and Frequency
Reduce the string length to 2/3 its open length and that is an interval of a fifth. That is a change
in frequency of D=293.665 Hz to B=493.88 Hz. Reduce the string length to 3/4 its open length
and that is a change in frequency from D=293.665 Hz to A=440 Hz, which is an interval of a
fourth. Cut the string length in half and you double the frequency. Which is an octave. We have!
!
!
!
Where is the golden ratio and is the golden ratio conjugate. And,…!
!
Thus, we see string length is inversely proportional to frequency. There are two equations for
string length, which we can examine by looking at a guitar (See figure 5). From the bridge of a
guitar to the fret, where the open string length is s we have:!
!
And!
!
Where s is the string length from the bridge to the nut and the nut is , fret 1 is , fret 2 is ,…!
is the distance from , and is the distance from , and so on…!
The 3/2 is an approximation to the the golden ratio (Phi).!
We can think of electron orbits in a hydrogen atom as frequencies related to length as well
because for a drop in orbit a photon is emitted that has a frequency associated with it. The
change in orbit is like the change in the length of a string. Since the n=3 orbit is a distance of
R3=4.761E-10m from the nucleus and R2=2.116E-10m we have R3/R2=2.25. But the Energy at
E3 is E3=-13.6eV/9, and at E2 it is -13.6eV/4. This is a ratio of 4/9=0.444 and 1/0.444==2.25,
thus there is an inverse relationship between length and frequency in the atom (See Fig. 5).!